Some of the following compounds show aromatic properties, and ...

Question

Some of the following compounds show aromatic properties, and others do not.

1. Predict which ones are likely to be aromatic, and explain why they are aromatic.

(a)

(b)

(c)

Accepted Answer

Hey everyone and welcome back in today's video, we are going to solve the following problem which compounds are likely to be aromatic, explain why they are aromatic. Now we are given three structures and before we begin solving our problem, we will just define two crates, syria for our metis city. We want to make sure that the structure is plainer cyclic I system and number two that the number of pi electrons would be four and plus two where an is an integer. This is defined by the Heracles rule. Okay, so we are ready to begin our problem, notice how each structure contains oxygen atoms. So before we can apply these two criteria, we have to understand whether or not the lone pair of oxygen is de localized. To do that. We need to recall the concept of resonance. First of all, let's redraw the structure, let's add oxygen tower wing. And I was looking at the first structure, we understand that indeed the lone pair on oxygen is de localized. Notice how we can use this lone pair and we can form a pi bond out of it because there's an adjacent pi bond that we can break. And actually we are not going to break it right because there's a positive charge. So we can actually shift the actual pi electrons within the ring over here to change the position of these pi electrons are basically de localize them as well at a different carbon atom. So this allows us to get our residence form and then we will be able to tell whether the structure is aromatic. So now we have our eye bon over here. This is our shifted pi bon. And then if we incorporate oxygen in our structure, let's not miss that. This be our oxygen. We understand that we use the lone pair in the formation of a new pi bon and this produces a positive charge and oxygen because now it has three bonds. And now let's analyze the structure. First of all, it's a planer cyclic by system. Yes, it is. Notice how we have our cyclic by system. We have pi electrons in a ring structure. And the reason why explainers because if you look at the carbon atoms, these are double bonded carbon atoms, right? And specifically double bonded carbon atoms are sp two hybridized. If they're sp two hybridized, you may recall that according to the V. S. E P R theory that would correspond to tribunal planner arrangement. So the structure so far is plainer thinking about this oxygen. If we count the number of electron regions around it, we may recall that there is still one lone pair remaining. So we have one single bond, one double bond and a lone pair. Three electron regions imply that oxygen is also sp two hybridized because all of the six atoms within our ring are sp two hybridized. The first criterion is satisfied. We can say that criteria. # one is satisfied because this is a plane or structure. And now let's count the number of pi electrons. We have three pi bonds three times two gives us six pi electrons. So if we have six by electrons, let's essentially equate that to our equation. And as long as we get an as an imager, we are good and we can say that the hiker's rule is satisfied. So if sex equals for m plus two, then if we subtract C from both sides for equals for em and it follows that and equals one. So this is an integer. And because this is an integer, the second rule is satisfied as well. So criteria number one and 2 are satisfied, meaning the first structure is aromatic. Now moving on to the 2nd because once again we have oxygen, we are going to draw the resonance structure of it. Let's incorporate the oxygen. Now, once again, we're going to think about the long pair of oxygen. And just as we saw it before, because there's an adjacent pi bond, we can de localize this lone pair of electrons. So it's actually de localized in the pi system. And we are going to shift the other pi bond over here to a different position because we're forming a pi bond here, we're going to shift this one. And because we're creating a pi bond at a different position, we have to break the adjacent one and we are going to essentially transfer our pi electrons in the form of a lone pair onto the top oxygen. We are now ready to draw our residents form. So that will be our six member ring. This is our original pi bond. This is the newly created by bond and this is another new pi bond. We have oxygen with a formal negative charge because there's only one bond. Now, it's really important to understand that we we have a positive charge in this oxygen. It has three bonds and we are going to add our metal group. So now criteria criteria number one, is it a cyclic planner pi system? If we look at our carbon atoms and oxygen in the ring, right, we essentially notice that these are our double bonded carbon atoms which are tribunal planner and similarly to what we saw before this oxygen is also sp two hybridized. So indeed because the structure contains only sp two hybridized atoms within the ring. Right. We're not considering the external substitutions. That implies the structure is indeed cyclic. Playing our pi system, it has these pi electrons. So we are ready to look at the second criterion, we can count the number of I bonds that we are observing. We have 33 times two. Once again gives us six pi electrons. And if there are six by electrons, we already know that the energizer that corresponds to six equals for employers to is one because it's an integer. That implies the structure follows the hot coals rule And therefore six by electrons would correspond to an aromatic structure. So going back because the two criteria are satisfied. This means the second structure is also aromatic. Moving on to the 3rd 1, we are once again going to think about the residence form. Okay, so this is our structure. And once again we are going to immediately think about the lone pair of oxygen because there's an adjacent pi bond, that means the lone pair is de localized because we can perform this shift. However, we are not going to think about this structure where we are creating a double bond in this position because we're going to transfer these pi electrons over here and we definitely don't want to have a lone pair on our ring. We want to make sure that we are only seeing pi bonds. So even though this is a valid resonance structure, that's perfectly fine, we are going to make sure that we are not breaking any of the pi bonds originally present in the ring instead to achieve all of our pi electrons within the ring. We can de localize this lone pair over here because we have an adjacent pipe bond on the external substitution. So we are going to break the adjacent pipe bond and transfer the electrons onto oxygen. This allows us to get our residence form which looks as follows. That would be our oxygen with a formal negative charge because now it has one bond, then we are drawing our ring and now this oxygen has three bonds, which implies it has a formal positive charge. Okay, so criteria number one, let's look at our sp two hybridized carbon atoms which are double bonded. And as we said before this oxygen has three electron regions. So it's also sp two hybridized, meaning the first grade here and is satisfied. This is a Lehner cyclic pi system. We have our pi electrons. This is a ring. All of the atoms within the ring. R. S P two hybridized. And as we said it before. If we check the high cultural 123, there are three pi bonds. Each bond has two electrons, six pi electrons. We already know that this implies the structure would follow the high cultural because once again recall that if six equals for employers to then an equals one. Right? So that's an integer. And because that's an integer, the second grade Syrian is satisfied as well, meaning the third structure is also aromatic. Let's add that to our answer. So it's aromatic as well. And let's label our final answers. So the first structures aromatic. The 2nd 1 is aromatic as well as the 3rd 1. Based on our answers, we may conclude that the correct answer to the multiple choice question is d However, if you have any questions, don't hesitate to leave a comment down below. Thank you for watching

Some of the following compounds show aromatic properties, and others do not.
1. Predict which ones are likely to be aromatic, and explain why they are aromatic.
(a)
(b)
(c)

Key Concepts

Aromaticity

Hückel's Rule

Conjugation

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